Question: In regular pentagon $ABCDE$, diagonal $AC$ is drawn, as shown. Given that each interior angle of a regular pentagon measures 108 degrees, what is the measure of angle $CAB$?

[asy]
size(4cm,4cm);
defaultpen(linewidth(1pt)+fontsize(10pt));

pair A,B,C,D,E;
A = (0,0);
B = dir(108);
C = B+dir(39);
D = C+dir(-39);
E = (1,0);

draw(A--B--C--D--E--cycle,linewidth(1));
draw(A--C,linewidth(1)+linetype("0 4"));

label("A",A,S);
label("B",B,W);
label("C",C,N);
label("D",D,E);
label("E",E,S);
label("$108^\circ$",B,E);;
[/asy]
Since $ABCDE$ is a regular pentagon, we know by symmetry that the measures of $\angle CAB$ and $\angle BCA$ are equal. We also know that the sum of the measures of the angles of $\triangle ABC$ equals $180$ degrees. Thus, if we let $x = $ the measure of $\angle CAB$ = the measure of $\angle BCA$, we have that $180 = 108 + x + x \Rightarrow 2x = 72 \Rightarrow x = 36$. The measure of angle $CAB$ is $\boxed{36}$ degrees.